How would you compute the following integrals? In=∫0π1−cos⁡nx1−cos⁡xdx Jn,m=∫0πxm(1−cos⁡nx)1−cos⁡xdx

Question

How would you compute the following integrals?  In=∫0π1−cos⁡nx1−cos⁡xdx  Jn,m=∫0πxm(1−cos⁡nx)1−cos⁡xdx

Carl Swisher · Accepted Answer

As for the first integral In=∫0π1−cos⁡nx1−cos⁡xdx=∫0πsin2(nx2)sin2(x2)dx=2 ∫0π20sin2(nx)sin2(x)dx=2∫0π2(sin⁡(nx)sin⁡(x))2dx= 2∫0π2(e∈x−e−∈xeix−e−ix)2dx=2 2∫0π2(∑k=0n−1(eix)2k+1−n)2dx=2 ∫0π2(∑k=0n−1(eix)2k+1−n)(∑l=0n−1(eix)2l+1−n)dx= ∫0π2(eix)2k+2l+2−2ndx= 2∑k=0n−1∑l=0n−1π2σ2k+2l+2−2n=π∑k=0n−1∑l=0n−1σ2k+2l+2−2n =πn As for the second... I&#039;ll think about it.

Kirsten Davis · Accepted Answer

The exponential generating function of Jm,n for fixed n is  Gn(t)=∑m=0∞Jm,ntmm≠∫0πetxsin⁡(nx)2sin⁡(x)2dx  We then have (sin⁡(nx)sin⁡(x))2=n+2∑j=1n−1(n−j)cos⁡(2jx) and  Gn(t)=(eπt−1)(nt+2∑j=1n−1(n−j)tt2+4j2)  For example, G3(t)=(eπt−1)(3t+4tt2+4+2tt2+16). Now  eπt−1=∑j=1∞πjtjj!=πt+π2t22+π3t36+…  3t+4tt2+4+2tt2+16=3t+…  ∑j=0∞(−1)j(4−j+16−j8)t2j+1=3t+(1+18)t−(14+116×8)t3+…  and so Jm,3 is m! times the coefficient of tm in the product of these.

karton · Accepted Answer

Another generating function approach is to set z=eix in ∑k=0k=0(it)mm!∫0πxm1−cos⁡(nx)1−cos⁡(x)dx=∫0πzt(zn−1z−1)2dzizn=1i∫0π(∑k=1−nn−1(n−|k|)zk+t−1)dz=1i∑k=1−nn−1(n−|k|)(−1)keπit−1k+t=∫1−eπitt−2it∑k=1n−1n−kk2−t2(1−(−1)keπit)

How would you compute the following integrals? I_n=\int_0^\pi\frac{1-\cos nx}{1-\cos x}dx J_{n,m}=\int_0^{\pi}\frac{x^m(1-\cos nx)}{1-\cos

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New Questions in Calculus 1